Problem: We know that $\frac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n-1}}+...$ for $x\in\left(-1,1\right)$. Using this fact, find the function that corresponds to the following series. $ x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+...+\frac{{{x}^{n}}}{n}+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\frac{-1}{{{\left( 1-x \right)}^{2}}}$ (Choice B) B $\frac{1}{{{\left( 1-x \right)}^{2}}}$ (Choice C) C $ -\arctan \left( 1-x \right)$ (Choice D) D $\arctan \left( 1-x \right)$ (Choice E) E $-\ln \left( 1-x \right)$ (Choice F) F $\ln \left( 1-x \right)$
First, note the following: $\int{\big(1+x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n-1}}+...\big)}dx=C+x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+...+\frac{{{x}^{n}}}{n}+...\,$ Using the $~u$ -substitution $~u=1-x~$ with $~du=-dx~$ gives us $\int{\frac{1}{1-x}}\text{ }dx=-\ln \left( 1-x \right)+D$ Then $ -\ln(1-x)+D=C+x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+...+\dfrac{{{x}^{n}}}{n}+...$ We subtract $D$ from both sides, getting a new constant ( $C-D$ ) on the right. Call that new constant $K$. $ -\ln(1-x)=K+x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+...+\dfrac{{{x}^{n}}}{n}+...$ Now let $x=0$ to see that $K=0$. Hence $ x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+...+\dfrac{{{x}^{n}}}{n}+...= -\ln(1-x)$